If an ammeter read 2A, another ammeter would probably also read 2A.
Keep in mind that every measurement involves error, and the equipment has a stated accuracy. If, for instance, the ammeter used had a 1% tolerance, than 2A actually means anywhere between 1.98A and 2.02A.
Put a known small resistance (maybe 1 ohm) in parallel to it. Then use the equation I=V/R (R being the known resistance, and V being the voltage readout) So when using a 1 ohm resistor, if the voltmeter reads 2.35V, it means I=2.35/1 = 2.35 A
2000ma is equal to 2 amps. Set you meter to a current range ABOVE 2 amps.
i=v/r can be used to it
Inductance = Magnetic Flux/Current = [ML2T-2A-1]/[A] = [ML2T-2A-2] So, Dimensional Formula of Inductance = [ML2T-2A-2]
An ammeter either directly, or via a current transformer, is used to measure current in a circuit.Current transformers (CTs) are necessary when either (1) you need to measure the large currents flowing in high-voltagesystems (a CT will isolate the HV system from the ammeter to which it is connected, as well as reduce the value of the HV current) or (2) you need to reduce the very large currents to be measured in medium-voltage systems.
The multiplicative inverse of -2a is 1/(-2a) = -1/2a
-2a + 5 = 1 -2a = -5 + 1 -2a = -4 a = -4/-2 a = 2
1/(2a)
3a-2a = 1
If I've read your question correctly, you need to subtract: a2 +2a -7 a2 -4a2 +5a2 -6 = 2a2 -6 Note, if x - y = z, then y = x - z; so: 2a2 -6 - (a2 -2a +1) = 2a2 -6 - a2 +2a -1 = a2 +2a -7
(2a + 1)(2a - 5)
2a + 1 = 3subtract 1 from both sides2a = 2divide both sides by 2a = 1
The voltage remains the same across the circuit as it is a parallel connection. So, the current across the upper half of the circuit where the ammeter is connected is calculated as I = V/R = 12.04 (total voltage)/12 (Resistance R1) = 1 A. Hence, the ammeter will read 1 A.
2a-3ab = -1
It is 2a - 16.
1
(2a+1)(a+1)